# The Laplace transform from scratch

Here's a natural way to derive the Laplace transform via taking the concept of a taylor series and making it continuous. We will, of course, start with a taylor series whose coefficients are given by $a(n)$

$\qquad \qquad \displaystyle\sum_0^{\infty} a(n) x^n \qquad x<1$

Then, make this continuous by replacing the sum by an integral

$\qquad \qquad \displaystyle\int_0^{\infty} a(n) x^n dn \qquad x<1$

We can change $x^n$ to $e^n \ln x$ and make the substitution $s =- \ln x$ to give us

$\qquad \qquad \displaystyle\int_0^{\infty} a(n) e^{-sn} dn \qquad s > 0$

Which is just the definition for the Laplace transform of $a(n)$, $\mathcal{L}(a)(s)$. Note by Levi Walker
1 year, 7 months ago

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Nice!

- 1 year, 3 months ago